You plan to take a trip to the moon. Since you do not have a traditional spacesh

Question

You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the earth with enough speed to make it to the moon. Some information that will help during this problem:

mearth = 5.9742 x 1024 kg
rearth = 6.3781 x 106 m
mmoon = 7.36 x 1022 kg
rmoon = 1.7374 x 106 m
dearth to moon = 3.844 x 108 m (center to center)
G = 6.67428 x 10-11 N-m2/kg2

1)

On your first attempt you leave the surface of the earth at v = 5534 m/s. How far from the center of the earth will you get?

m

2)

Since that is not far enough, you consult a friend who calculates (correctly) the minimum speed needed as vmin = 11068 m/s. If you leave the surface of the earth at this speed, how fast will you be moving at the surface of the moon? Hint carefully write out an expression for the potential and kinetic energy of the ship on the surface of earth, and on the surface of moon. Be sure to include the gravitational potential energy of the earth even when the ship is at the surface of the moon!

m/s

3)

Which of the following would change the minimum velocity needed to make it to the moon?

the mass of the earth

the radius of the earth

the mass of the spaceship

(Survey Question)

Explanation / Answer

Mass of the Earth is, ME = 5.9742 * 1024 kg

Radius of the Earth is, RE = 6.3781 * 106 m

Mass of the moon is, Mm = 7.36 * 1022 kg

Radius of the moon is, Rm = 1.7374 * 106 m

Distance from the Earth to the moon (center to center) is, dE-m = 3.844 * 108 m

Universal gravitational constant is, G = 6.67428 * 10-11 N-m2/kg2

1)

Velocity of the vehicle is, v = 5534 m/s

Let us suppose that the vehicle reaches a distance R from the center of the earth. Then, the kinetic energy of the vehicle should be equal to the change in the potential energy of the vehicle.

At the surface of the earth, gravitational potential energy of the vehicle is,

Us = -GME m / RE

And at a distance R,

UR = -GME m / R

Kinetic energy of the vehicle is, T = (1/2) mv2

Thus,

(1/2) mv2 = UR - Us = GMEm [1/RE - 1/R]

0.5 * 5534 * 5534 = 6.67428 * 10-11 * 5.9742 * 1024 * [1/6.3781 * 106 - 1/R]

[0.156786504 * 10-6 - 1/R] = 384029.099710182 * 10-13

1/R = 0.156786504 * 10-6 - 384029.099710182 * 10-13

R = 8447116.413782808 m = 8447116. 41 m

2)

vmin = 11068 m/s.

Let us assume that v is the velocity of the ship on the surface of the moon.

Total energy of the spaceship on the surface of the earth is,

Ee = (1/2) mvmin2 - GMEm/RE - GMmm/(DE-m - RE)

Total energy of the spaceship on the surface of the moon is,

Em = (1/2) mv2 - GMmm/Rm - GMEm/(DE-m - Rm)

From the law of conservation of energy,

Ee = Em

(1/2) mvmin2 - GMEm/RE - GMmm/(DE-m - RE) = (1/2) mv2 - GMmm/Rm - GMEm/(DE-m - Rm)

(1/2)vmin2 - GME/RE - GMm/(DE-m - RE) = (1/2)v2 - GMm/Rm - GME/(DE-m - Rm)

0.5 * 11068 * 11068 - 6.67428 * 10-11 * [5.9742 * 1024 / 6.3781 * 106 + 7.36 * 1022 / (3.844 * 108 - 6.3781 * 106)] = 0.5 * v2 - 6.67428 * 10-11 * [7.36 * 1022 / 1.7374 * 106 + 5.9742 * 1024 / (3.844 * 108 - 1.7374 * 106)]

61250312 - [0.156786504 * 10-6 - 0.002613268 * 10-6] * 6.67428 * 10-11 * 5.9742 * 1024 - [0.002645349 * 10-6 - 0.575572695 * 10-6] * 6.67428 * 10-11 * 7.36 * 1022 = 0.5 * v2

61250312 -6.147423994 * 107 + 28.143738598 * 105 = 0.5 * v2

64064685.8598 - 61474239.94 = 0.5 * v2

0.5 * v2 = 2590445.9198

v = 2276.157252828 m/s = 2276.16 m/s

3)

It can be seen from part 1), that the minimum velocity needed to make it to the sun would change if either or both mass of the earth and the radius of the earth change.

Change in the mass of the spaceship will not have any impact on the minimum velocity required to make it to the moon.

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